Getting the file name in a package
- To: mathgroup at smc.vnet.net
- Subject: [mg52885] Getting the file name in a package
- From: Geronimo <kalymereauKILLSP at Myahoo.fr>
- Date: Thu, 16 Dec 2004 03:40:18 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Hi I have a package (.m) that produces some output in an external file. As the computation is quite long, I would like to avoid it if the output file is more recent than the last modification of the package (since the output only depends on what is in the package). So I need to extract the file name of the package within itself (and then test the date with FileDate). As the context name is related to the file name, I use: BeginPackage["TestPack`"] (blabla) Begin["`Private`"] currentFileName = StringDrop[Context[], -9] <> ".m" (blabla) End[] EndPackage Since within the Private context the Context[] command gives TestPack`Private` and `Private` is 9 characters long, it works: currentFileName is equal to "TestPack.m". However I find this solution quite cumbersome, is there a more obvious way to do that? Thank you Geronimo